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anonymous
 4 years ago
Rationalize the denominator: cubic root of 2y^4/6x^4
anonymous
 4 years ago
Rationalize the denominator: cubic root of 2y^4/6x^4

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt[3]{2y ^{4}/6x ^{4}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My instinct is to go down instead of making it \[x ^{64}\] because that would look dumb, unless it's correct but I've never had to decrease... do I just multiply by a negative exponent?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0=cube root (2y^4)/cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/cube root(6x^4)*cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/6x^4 basically separate top and bottom roots multiply by conjugates fix ugly maths

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.0cuberoot(y/3x)^4 =(y/3x)cuberoot(y/3x)=(ycuberoot3xy)/3x

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.02/6 simplifie by 2 and get 1/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm, well that is handy, didn't know oyu could simplify it but what would the correct conjugate be?

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.01/sqrt3 =(sqrt3)/3 like example

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No actually, I'm confused... can you just tell me what I need to multiply by to get rid of the radical in the denominator?

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.01/sqrt3 so when there is radical in denominator multiplie numerator and denominator by conjugate so in this case by sqrt3 and in this way you get in denominator 3 and in numerator sqrt3

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.0because sqrt3 * sqrt3 =3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know... but what would the conjugate be. I know how to do all this already i get it I just need to know if the conjugate I am using is correct...

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.0your answer not is right

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.0check my answer step by step

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0we have to simplify that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok look, the denominator is cubic root of 6x^4 .... so what do I multiply it by to take a perfect cubic root?

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0If sat is here, then there was no need of calling me! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then you will get in the denominator a perfect cube, namely \[\sqrt[3]{216x^6}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0He just got here thanks anyway

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you have \[\sqrt[3]{6x^4}\] so to make it a perfect cube, you need the exponent on the x to be 6, and you have to multiply by 6^2=36 to make the constnant a cube

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait why does the exponent for x have to be 6?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because you have a cube root so you need a number divisible by 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i mean an exponent divisible by 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt[3]{216x^6}=6x^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So that would come out as \[6x ^{2}\] ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah thank you that is ALL I needed, I guess nobody understood what I meant. Thank you very much
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